Question

1. a bird is flying 77.0 meters above the ground, holding a stick with its feet. if the bird were to drop the stick, how long would it take for the stick to reach the ground below? 3.48 s 3.96 s 1.96 s 2.45 s

Explanation:

Step1: Identify the kinematic equation

The kinematic equation for vertical motion under gravity (initial velocity \(u = 0\)) is \(h=ut+\frac{1}{2}gt^{2}\). Since \(u = 0\), the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(h\) is the height (\(h = 77.0\space m\)), \(g\) is the acceleration due to gravity (\(g=9.8\space m/s^{2}\)), and \(t\) is the time.

Step2: Solve for \(t\)

From \(h=\frac{1}{2}gt^{2}\), we can express \(t\) as \(t=\sqrt{\frac{2h}{g}}\).
Substitute \(h = 77.0\space m\) and \(g = 9.8\space m/s^{2}\) into the formula:

$$t=\sqrt{\frac{2\times77.0}{9.8}}=\sqrt{\frac{154}{9.8}}=\sqrt{15.714}\approx3.96\space s$$

Answer:

3.96 s